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Am I Ever Stuck

  1. Aug 29, 2006 #1
    The starship enterprise returns from warp drive to ordinary space with a forward speed of 50 km/s. To the crew’s shocking surprise, a Klingon ship is 100 km directly ahead, traveling in the same direction at a measly 20 km/s. Without evasive action, the Enterprise will overtake and collide with the Klingons in just slightly over 3.0 s. The Enterprise’s computers react instantly to brake the ship. What acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship? Assume the acceleration is constant.

    By the way, if it seems familiar it is on this thread, he must have had same textbook, https://www.physicsforums.com/archive/index.php/t-91330.html roffle

    so d(klingon)= vt + 100
    d(enterprise)= .5(vf - vi)t + vi(t) (I got from area under graph)

    they are equal I thought when they stopped so I combined them:
    vk(t) + 100 = .5(vf-vi)t + vi(t)
    vk(t) + 100 = .5vi(t)
    20t + 100 = 25t
    t = 20 s

    since I hope I have proven my attempt I ended up getting an accerlation of -2500 m/s^2

    I am sort of confident I didn't make a mathematical error since I checked it a lot, so I assume I made a conceptual error. Answer in back is -4500 m/s^2. I must have underestimated the question?

    Thanks for any help you might have to direct me and I am sorry I do not know how to make the equations fancy.
     
  2. jcsd
  3. Aug 29, 2006 #2

    chroot

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    Gold Member

    I have no idea how you arrived at the equation

    d(enterprise)= .5(vf - vi)t + vi(t)

    The equation for the position of any body undergoing acceleration is:

    [itex]s_{enterprise}(t) = s_{0, enterprise} + v_{0, enterprise} t + \frac{1}{2} a_{enterprise} t^2[/itex]

    The equation for the klingon ship is simpler, because it is not undergoing any acceleration:

    [itex]s_{klingon}(t) = s_{0, klingon} + v_{0, klingon} t[/itex]

    The initial velocities are known, as are the initial positions.

    If the two ships barely miss each other, then they can be considered to be essentially the same position at some time t.

    - Warren
     
  4. Aug 29, 2006 #3
    okay, I'll tell you how I got at it:
    d= d1 (which is 0) + vi(t) + .5at^2
    so I did a= -vi/t in there
    and that's how I got it.

    my whole d(enterprise) = .5vi(t)

    http://dev.physicslab.org/Document....=Kinematics_DerivationKinematicsEquations.xml

    The end of equation 3. Yeah, I looked too. But I wrote from area under the graph because I sort of did it graphically... Maybe that threw you off.

    I am guessing you didn't look at any of my work because it is not fancy. ;'( tear.

    Aaand I'm camping here.

    Edit: In an attempt to make myself clearer without using fancy equations (I'm hopeless)
    - d(enterprise) = .5(vo + vi)t from website.
    - d(klingon) = vk(t) + 100
    - I assumed distance to be the same at end of question.
    - .5(vo +vi)t = vk(t) + 100
    - 25t = 20t + 100
    t = 20 s
    I put that into
    a = (vf - vi)/t
    to get: -2,500 m/s^2

    WHICH IS WROOOOOONG.
    I hope this is more clear. I am sorry to be wasting the authority figures of physics times.
     
    Last edited: Aug 29, 2006
  5. Aug 29, 2006 #4

    mezarashi

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    Homework Helper

    I lost you on how you arrived at some of the equations, but this question can be much simplified if you use one trick, that is, the concept of relative velocities. Or else I can see that you have a pretty complicated situation where Enterprise's braking distance increases or decreases based on its planned decceleration due to Klington's moving.

    Consider that Klingon is non-moving, and that Enterprise is moving towards it at 50-20 = 30km/s. Enterprise is 100 km away. At what acceleration does Enterprise need to stop so that it just doesn't collide?

    The following equation should come to mind:
    [tex]v_f^2 = v_i^2 + 2ad[/tex]

    I hope that helps ;)
     
    Last edited: Aug 29, 2006
  6. Aug 29, 2006 #5
    Thanks much!!
    I also get what I was doing wrong. I was for some reason assuming the enterprise stopped so vf = 0. OH SNAP!! I assumed vf for enterprise is equal to velocity of klingon at the end and it turned out right.
     
    Last edited: Aug 29, 2006
  7. Aug 29, 2006 #6
    Yes, in this situation I would imagine the klingon is going 0km/s and the enterprise doing 30km/s this avoids any unnessisary confusion (and obviously gives the same result if they're going 20km/s and 50km/s respectively) .
     
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